3N^'+4tN^6=0

Lösung

3 N^{^{'}} + 4 t N^{6} = 0

N (t) = - 5 \frac{3}{5 ( - 2 t ^{2} + c _{1} )}

Schritte zur Lösung

3 N^{'} (t) + 4 tN (t)^{6} = 0

Solve separable ODE:

N (t) = - 5 \frac{3}{5 ( - 2 t ^{2} + c _{1} )}

N (t) = - 5 \frac{3}{5 ( - 2 t ^{2} + c _{1} )}

x^{^{'}} = x^{2} + x + 1

(y x^{2} - y) d y = arctan (x) d x

y^{^{'}} = (t - 1) (2 y + 1)

y^{^{'}} = - y \cdot cos (x)

y^{2} sec (5 x) \frac{d y}{d x} = 1, y (0) = 0